Senior Maths Challenge question

123ohrid · #1 $Old$ November 5th, 2009, 04:32 PM

Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being difficult. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

Find a+b+c+d

Anyone got any idea how to do this?

Danny · #2 $Old$ November 5th, 2009, 04:57 PM

Quote:

Originally Posted by 123ohrid $View Post$

Hello lads. So I did my senior maths challenge today, and there was a question which struck me as being the sort of WTF questions. I even asked my maths teacher and he couldn't solve it as fast as he could. So who can run me through how you would solve the following:

abcd + abc + bcd + acd + abd + ab + bc + cd + ad + ac + bd+ a + b + c + d = 2009

Find a+b+c+d

Anyone got any idea how to do this?

Others may prove me wrong but I don't think there's a unique answer to this question. You'll notice that adding 1 to both sides gives

$(a+1)(b+1)(c+1)(d+1) = 2010.$

Some solutions

$a = 1, \;b = 1004, \; c = 0, \;d = 0,$

$a = 2, \;b = 669, \; c = 0, \;d = 0,$

$a = 4, \;b = 401, \; c = 0, \;d = 0,$

$a = 5, \;b = 334, \; c = 0, \;d = 0,$

$a = 9, \;b = 200, \; c = 0, \;d = 0.$

We see that $a+b+c+d$ changes in all these cases.

Jameson · #3 $Old$ November 5th, 2009, 05:14 PM

I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.

Danny · #4 $Old$ November 5th, 2009, 05:46 PM

Quote:

Originally Posted by Jameson $View Post$

I bet the question said none can equal 0. If so, 2010 breaks down into 4 prime factors which furthers this conclusion - 2,3,5,67.

Nice observation!

123ohrid · #5 $Old$ November 6th, 2009, 12:13 AM

Yeah it sayd that a b c and d are positive integers.

Also this might help, the answer is one of the following:

A)73 B) 75 C) 77 D) 79 E) 81..

So which one would it be?

I'm guessing 77 because it's the sum of all your primes? But can you try to explain why?

Defunkt · #6 $Old$ November 6th, 2009, 01:40 AM

$(a+1)(b+1)(c+1)(d+1) = 2010$

2010 has exactly 4 prime factors -- 2, 3, 5, 67 (this means that $2010 = 2\cdot 3 \cdot 5 \cdot 67$ . Convince yourself that you cannot write 2010 as another product of 4 different factors.), so $(a+1) + (b+1) + (c+1) + (d+1) = 2 + 3 + 5 + 67 \Rightarrow a + b + c + d = 73$